Problem 22 - Calculus Early Transcendentals

Question

a. Find the first four nonzero terms of the Taylor series for the given function centered at \(a\) b. Write the power series using summation notation. $$f(x)=\cos x, a=\pi$$

Step-by-Step Solution

Answer

The first four nonzero terms in the Taylor series of the function \(f(x) = \cos x\) centered at \(a = \pi\) are: $$f(x) = -1 + \dfrac{(x-\pi)^4}{4!} + ...$$ In summation notation, the power series for the function is: $$f(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!} (x-\pi)^{2n}$$

Step 1: 1. Find the first few derivatives of the function

To find the Taylor series, we first need to find the derivatives of the function. $$f(x) = \cos x$$ $$f'(x) = -\sin x$$ $$f''(x) = -\cos x$$ $$f^{(3)}(x) = \sin x$$ $$f^{(4)}(x) = \cos x$$

Step 2: 2. Evaluate the derivatives at \(a=\pi\)

Now, we need to evaluate the derivatives at the point \(a=\pi\). $$f(\pi) = \cos \pi = -1$$ $$f'(\pi) = -\sin \pi = 0$$ $$f''(\pi) = -\cos \pi = 1$$ $$f^{(3)}(\pi) = \sin \pi = 0$$ $$f^{(4)}(\pi) = \cos \pi = -1$$

Step 3: 3. Use the Taylor series formula to construct the series' terms

The Taylor series formula is given by: $$f(x) = \sum_{n=0}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x-a)^n$$ Using the values we've found at step 2, we can now write the first four nonzero terms of the Taylor series for \(f(x) = \cos x\) centered at \(a=\pi\): $$f(x) = -1 + \dfrac{0}{2!}(x-\pi)^2 + \dfrac{(-1)}{4!}(x-\pi)^4 + ...$$ Since the second term is zero, we only display the first, second, third, and fourth terms of the series: $$f(x) = -1 + \dfrac{(x-\pi)^4}{4!} + ...$$

Step 4: 4. Construct the power series using summation notation

Now, we can write the power series in summation notation. For \(f(x)=\cos x\), we can see that even derivatives are nonzero and odd derivatives are zero. Thus, the summation is over even n-values only: $$f(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!} (x-\pi)^{2n}$$ So, the first four nonzero terms in the Taylor series of the given function centered at \(a=\pi\) are: $$f(x) = -1 + \dfrac{(x-\pi)^4}{4!} + ...$$ And in summation notation, the power series for the function is: $$f(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n)!} (x-\pi)^{2n}$$